exec命令不适用于添加的变量，但是可以在不添加变量的情况下使用

Question

I'm trying to execute python file made by CaryKH called jumpcutter.py using child process

This is a working function, where the exec command does work:

function executePython(){

  const { exec } = require('child_process');
  const path = require('path');
  var input = document.getElementById('input').value
  var output = document.getElementById('output').value
  var silent = document.getElementById('silent').value
  var sounded = document.getElementById('sounded').value
  var margin = document.getElementById('margin').value
  alert(input)

  exec('python jumpcutter.py --input_file wow.mp4 --output_file wowcut.mp4 --silent_speed 999999 --sounded_speed 1 --frame_margin 1')
}

However, when I do this:

exec('python jumpcutter.py --input_file ' +input ' --output_file ' +output ' --silent_speed ' +silent ' --sounded_speed ' +sounded ' --frame_margin ' +margin)

the code doesn't work entirely, so even the alert no longer works, even though it did prior. I have already tried to store the console command in a variable called text but to no prevail.

Thank you in advance for any assistance :)

Answer 1

the string concatenate is wrong

to concatenate multiple strings into one the syntax is:

string1 + string2 + string3 + string4 + ....

so your code should be like this:

exec('python jumpcutter.py --input_file ' + input +' --output_file ' +output+ ' --silent_speed ' +silent+ ' --sounded_speed ' +sounded+ ' --frame_margin ' +margin)